Two people are throwing a ball from shoulder height $S$ in a hallway with ceiling $H$ meters high. The ball is thrown with velocity $v$. Show that the maximum separation is $d=4\sqrt{(H-S)v^2/2g - (H-S)^2}$ if $v^2 \ge 4g(H-S)$ and $d = v^2/g$ if $v^2 \le 4g(H-S)$.
I am able to prove both these results but I have no understanding as to what is actually happening with these results.
I used mamimum height is $H-S = \frac{v^2\sin^2{a}}{2g}$ metres, and range $d=\frac{v^2}{g}\sin{2a}$.
For the $d = v^2/g$ case, it "works" if you just let $a = 45$ degrees. I don't know how the condition $v^2 \le 4g(H-S)$. comes into play though. Same with the other case. I just substitute in $H-S = \frac{v^2\sin^2{a}}{2g}$ to prove $d=4\sqrt{(H-S)v^2/2g - (H-S)^2}$ if $v^2 \ge 4g(H-S)$ but what is the difference between these cases?
Note that the second case can be expressed as $H\geq v^2/(4g)+S$, i.e., it's the case of the ceiling being sufficiently high. So I'd interpret this as: If the ceiling is high enough, then the maximum distance is just the usual maximum range. But if the ceiling is too low, then you'll need to angle the shot downwards: this will reduce the range, but it'll also lower the peak height.
Some more details: A quick calculation shows that the maximum height achieved is $y_{max}(\theta)=S+\dfrac{v^2}{2g}\sin(\theta)$, while the range (since the ball start and ends at the same height) is $R(\theta)=\dfrac{v^2}{2g}\sin(2\theta)$. Hence we're looking to maximize $R(\theta)$, subject to the constraint $y_{max}(\theta)\leq H$. Thinking about the behavior of $\sin(2\theta)$ and $\sin(\theta)$ on the range of angles $\theta\in (0,\pi/2)$, we obtain the story I sketched above.